Question

Equation of a tangent to the hyperbola \[5{{x}^{2}}-{{y}^{2}}=5\] and which passes through an external point (2, 8) is
A. 3x-y+2=0
B. 3x+y-14=0
C. 23x-3y-22=0
D. 5x-4y+22=0

Answer 1

Hint:Differentiate the equation of hyperbola and find the slope of tangent of hyperbola, where \[m=\dfrac{dy}{dx}\]. Then find the equation of tangent by using slope. Substitute the point (2, 8) in the equation of tangent to get the required equation.

Complete step-by-step answer:
Given us the equation of the hyperbola\[\Rightarrow 5{{x}^{2}}-{{y}^{2}}=5-(1)\]
The slope of tangent of the hyperbola is denoted by m.
\[\therefore m=\dfrac{dy}{dx}\], which is equal to the differential.
Now differentiate equation (1) with respect to ‘x’.
\[5{{x}^{2}}-{{y}^{2}}=5\]
\[\begin{align}
  & \Rightarrow 10x-2y.\dfrac{dy}{dx}=0 \\
 & \therefore +2y\dfrac{dy}{dx}=10x \\
 & \therefore \dfrac{dy}{dx}=\dfrac{10x}{2y} \\
 & \dfrac{dy}{dx}=\dfrac{5x}{y} \\
\end{align}\]
\[\therefore \]The slope of tangent of the hyperbola, \[m=\dfrac{dy}{dx}\]
\[m={{\left( \dfrac{dy}{dx} \right)}_{\left( 2,8 \right)}}\]as the equation of tangent to the hyperbola passes through the point (2, 8).
\[\begin{align}
  & \therefore m={{\left( \dfrac{dy}{dx} \right)}_{\left( 2,8 \right)}} \\
 & m={{\left( \dfrac{5x}{y} \right)}_{\left( 2,8 \right)}}=\dfrac{5\times 2}{8}=\dfrac{10}{8}=\dfrac{5}{4} \\
 & \therefore m=\dfrac{5}{4} \\
\end{align}\]
Equation of tangent is given by the formula.
\[\left( y-{{y}_{1}} \right)=m\left( x-{{x}_{1}} \right)\]
\[\left( {{x}_{1}},{{y}_{1}} \right)=\left( 2,8 \right)\]and \[m=\dfrac{5}{4}\]
\[\Rightarrow \left( y-8 \right)=\dfrac{5}{4}\left( x-2 \right)\]
Cross multiplying and simplify it,
\[4\left( y-8 \right)=5\left( x-2 \right)\]
Open the brackets,
\[\begin{align}
  & 4y-32=5x-10 \\
 & \Rightarrow 5x-4y+32-10=0 \\
 & 5x-4y+22=0 \\
\end{align}\]
Hence, the correct option is (D).

Note: The slope of tangent is given by \[m=\dfrac{dy}{dx}\] by differentiating the equation of parabola. So, it's important to find the derivative of the equation of hyperbola.Students should remember equation of tangent given by \[\left( y-{{y}_{1}} \right)=m\left( x-{{x}_{1}} \right)\]